cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A316566 Triangle read by rows: T(n,k) is the number of elements of the group GL(2, Z(n)) with order k, 1 <= k <= A316565(n).

Original entry on oeis.org

1, 1, 3, 2, 1, 13, 8, 6, 0, 8, 0, 12, 1, 27, 8, 36, 0, 24, 1, 31, 20, 152, 24, 20, 0, 40, 0, 24, 0, 40, 0, 0, 0, 0, 0, 0, 0, 48, 0, 0, 0, 80, 1, 55, 26, 24, 0, 98, 0, 48, 0, 0, 0, 12, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 24, 1, 57, 170, 42, 0, 618, 48, 84, 0, 0, 0, 84
Offset: 1

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Author

Andrew Howroyd, Jul 06 2018

Keywords

Comments

For coprime p,q the group GL(p*q, Z(n)) is isomorphic to the direct product of the two groups GL(p, Z(n)) and GL(q, Z(n)).

Examples

			Triangle begins:
  1
  1, 3, 2
  1, 13, 8, 6, 0, 8, 0, 12
  1, 27, 8, 36, 0, 24
  1, 31, 20, 152, 24, 20, 0, 40, 0, 24, 0, 40, 0, 0, 0, 0, 0, 0, 0, 48, 0, 0, 0, 80
  1, 55, 26, 24, 0, 98, 0, 48, 0, 0, 0, 12, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 24
  ...
		

Crossrefs

Row sums are A000252.
Column 2 is A066947.

Programs

  • PARI
    MatOrder(M)={my(id=matid(#M), k=1, N=M); while(N<>id, k++;N=N*M); k}
    row(n)={my(L=List()); for(a=0, n-1, for(b=0, n-1, for(c=0, n-1, for(d=0, n-1, my(M=Mod([a, b; c, d], n)); if(gcd(lift(matdet(M)), n)==1, my(t=MatOrder(M)); while(#L
    				

Formula

T(p*q,k) = Sum_{i>0, j>0, k=lcm(i, j)} T(p, i)*T(q, j) for gcd(p, q)=1.
T(n,k) = Sum_{d|k} mu(d/k) * A316584(n,k).

A316563 Maximum order of an element in the special linear group SL(2, Z(n)).

Original entry on oeis.org

1, 3, 6, 6, 10, 12, 14, 8, 18, 30, 22, 12, 26, 42, 30, 16, 34, 18, 38, 30, 42, 66, 46, 24, 50, 78, 54, 42, 58, 60, 62, 32, 66, 102, 70, 36, 74, 114, 78, 40, 82, 84, 86, 66, 90, 138, 94, 48, 98, 150, 102, 78, 106, 54, 110, 56, 114, 174, 118, 60, 122, 186, 126
Offset: 1

Views

Author

Andrew Howroyd, Jul 06 2018

Keywords

Crossrefs

Row lengths of A316564.

Programs

  • GAP
    Concatenation([1], List([2..15], n->Maximum(List(SL(2, Integers mod n), Order))));
    
  • PARI
    MatOrder(M)={my(id=matid(#M), k=1, N=M); while(N<>id, k++;N=N*M); k}
    a(n)={my(m=0); for(a=0, n-1, for(b=0, n-1, for(c=0, n-1, for(d=0, n-1, my(M=Mod([a, b; c, d], n)); if(matdet(M)==1, m=max(m, MatOrder(M))))))); m}

A316560 Number of cyclic subgroups of the group GL(2, Z(n)), counting conjugates as distinct.

Original entry on oeis.org

1, 5, 28, 62, 176, 148, 610, 696, 1252, 920, 2296, 1972, 4874, 3523, 6040, 6320, 8136, 7348, 14984, 13568, 22124, 11920, 17396, 23952, 29846, 28172, 38044, 47656, 47282, 32908, 75036, 53520, 71768, 42312, 145852, 99892, 123524, 88456, 187036, 179200, 152290
Offset: 1

Views

Author

Andrew Howroyd, Jul 06 2018

Keywords

Crossrefs

Programs

  • GAP
    Concatenation([1], List([2..7], n->Sum( Filtered( ConjugacyClassesSubgroups( GL(2, Integers mod n)), x->IsCyclic( Representative(x))), Size)));
    
  • PARI
    MatOrder(M)={my(id=matid(#M), k=1, N=M); while(N<>id, k++;N=N*M); k}
    a(n)={sum(a=0, n-1, sum(b=0, n-1, sum(c=0, n-1, sum(d=0, n-1, my(M=Mod([a, b; c, d], n)); if(gcd(lift(matdet(M)), n)==1, 1/eulerphi(MatOrder(M)))))))}

Formula

a(n) = Sum_{k=1..A316565(n)} 1/phi(A316566(n,k)).

A327568 Exponent of the group GL(2, Z_n).

Original entry on oeis.org

1, 6, 24, 12, 120, 24, 336, 24, 72, 120, 1320, 24, 2184, 336, 120, 48, 4896, 72, 6840, 120, 336, 1320, 12144, 24, 600, 2184, 216, 336, 24360, 120, 29760, 96, 1320, 4896, 1680, 72, 50616, 6840, 2184, 120, 68880, 336, 79464, 1320, 360, 12144
Offset: 1

Views

Author

Jianing Song, Sep 17 2019

Keywords

Comments

The exponent of a finite group G is the least positive integer k such that x^k = e for all x in G, where e is the identity of the group. That is to say, the exponent of a finite group G is the LCM of the orders of elements in G. Of course, the exponent divides the order of the group.

Examples

			GL(2, Z_2) is isomorphic to S_3, which has 1 identity element, 3 elements with order 2 and 2 elements with order 3, so a(2) = lcm(1, 2, 3) = 6.
		

Crossrefs

Programs

  • PARI
    MatOrder(M)={my(id=matid(#M), k=1, N=M); while(N<>id, k++; N=N*M); k}
    a(n)={my(m=1); for(a=0, n-1, for(b=0, n-1, for(c=0, n-1, for(d=0, n-1, my(M=Mod([a, b; c, d], n)); if(gcd(lift(matdet(M)), n)==1, m=lcm(m, MatOrder(M))))))); m} \\ Following Andrew Howroyd's program for A316565

Formula

If gcd(m, n) = 1 then a(m*n) = lcm(a(m), a(n)).
Conjecture: a(p^e) = (p^2-1)*p^e for primes p. If this is true, then 24 divides a(n) for n > 2.
Showing 1-4 of 4 results.